Structure of Atom Test

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Structure of Atom Test - 72

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Weekly Quiz Competition

Question 1
1 / -0

An ionic atom equivalent to hydrogen atom has wavelength equal to $$\dfrac{1}{4}$$ of the wavelengths of hydrogen lines. The ion will be

A
$$He^+$$

B
$$Li^{++}$$

C
$$Ne^{9+}$$

D
$$Na^{10+}$$

Solution

$$\text{By Rydberg's formula ,we have -}$$
$$\begin{array}{l}R_{H}=\text { Rydberg constant } \\\lambda=\text { wavelength } \\Z=\text { atomic num. }\end{array}$$

$$\begin{array}{l}\text { * For a particular member of spectral line }\\\text { we have - } \\\qquad \begin{aligned}\frac{1}{\lambda} &\alpha z^{2} \quad \Rightarrow\left[\lambda \alpha \frac{1}{z^{2}}\right] \\& \Rightarrow \lambda=\frac{k}{z^{2}} \text { (let) }-(1) \\\Rightarrow & \lambda_{H}=k\end{aligned}\end{array}$$

$$\begin{array}{l}\text { For any other ionic atom for which } \\\text { wavelength }=\frac{\lambda_{H}}{4}=\frac{k}{4}-(2) \\\text { from eq (1) and (2) } \\\text { we get } \frac{k}{z^{2}}=\frac{k}{4}\\\Rightarrow z^{2}=4 \\\Rightarrow z=2\quad\left(\mathrm{He}^{+}\right) \text {atom. } \\\text { Hence, option (A) is correct. }\end{array}$$
Question 2
1 / -0

The third line of Balmer series of an ion equivalent to hydrogen atom has wavelength of $$108.5nm$$. the ground state energy of an electron of this ion will be

A
$$3.4eV$$

B
$$13.6eV$$

C
$$54.4eV$$

D
$$122.4eV$$

Solution

For third of Balmer series $$n_1=2,\ n_2=5$$

$$\therefore \dfrac{1}{\lambda}=RZ^2 \left[\dfrac{1}{n_1^2}-\dfrac{1}{n_2^2} \right]$$

gives $$Z^2=\dfrac{n_1^2 n_2^2}{(n_2^2-n_1^2) \lambda R}$$

On putting values $$Z=2$$

From $$E=-\dfrac{13.6Z^2}{n^2}=\dfrac{-13.6(2)^2}{(1)^2}=-54.4\ eV$$
Question 3
1 / -0

The frequency of the characterstic X ray of $$K_a$$ line of metal targent 'M' is $$2500 cm^{-1}$$ and the graph between $$\sqrt { v } $$ Vs 'z' is as follows, then atomic number of M is?

A
$$49$$

B
$$50$$

C
$$51$$

D
$$25$$

Solution
Question 4
1 / -0

The mass of two moles of nitrogen atom is_____

A
$$28\space g$$

B
$$14\space g$$

C
$$7\space g$$

D
$$20\space g$$
Question 5
1 / -0

Angular momentum in second Bohr orbit of H-atom is $$x$$. Then find out angular momentum in 1st excited state of $${Li}^{+2}$$ ion:

A
$$3x$$

B
$$9x$$

C
$$\cfrac{x}{2}$$

D
$$x$$

Solution

$$ \begin{array}{l} \text { we know, } \\ \text { Angular momentum }=\frac{n h}{2 \pi} \end{array} $$
$$ \begin{array}{l} \text { Given, } \\ \text { For } n=2 \\ \text { Angular momentum = } x \\ \qquad x=\frac{2 h}{2 \pi} --(i) \end{array} $$
$$ \text { for } 1^{\text {st }} \text { excited state means } n=2 $$
$$ \begin{aligned} & A \cdot M=\frac{2 h}{2 \pi} \quad-(11) \\ \therefore \quad &(i)=(i i) \end{aligned} $$
$$ \text { Angular mementum }= x $$
Question 6
1 / -0

If one were to apply Bohr's model to a particle of mass 'm' and charge 'q' moving in a plane under the influence of a magnetic field 'B', the energy of the charged particles in the $$n^{th}$$ level will be:

A
n$$(\dfrac{hqB}{8\pi m})$$

B
n$$(\dfrac{hqB}{4\pi m})$$

C
n$$(\dfrac{hqB}{2\pi m})$$

D
n$$(\dfrac{hqB}{\pi m})$$

Solution

According to given data

We know that,

$$mvr=\dfrac{nh}{2\pi }$$
Or,

$$qvB=\dfrac{m{{v}^{2}}}{r}$$

So we find,

$$qB=\dfrac{mv}{r}$$

$$qB\left( \dfrac{nh}{2\pi mv} \right)=mv$$

$$ \left( \dfrac{1}{2}m{{v}^{2}} \right)=\dfrac{nhqB}{4\pi m} $$

$$ E=n\left( \dfrac{hqB}{4\pi m} \right) $$

This is the required solution.
Question 7
1 / -0

Assuming Bohr's model for $$ Li^{++} $$ atom, the first excitatio energy of ground state of $$ Li^{++} $$ atom is

A
10.2 eV

B
91.8 eV

C
13.6 eV

D
3.4 eV

Solution

For $$Li^{++}$$ model,
$$2=3$$
In first excitation of ground state,
$$n_1=1$$ and $$n_2=2$$
In Bohr's model,
$$\Delta E=(13.6ev)2^2\left[\dfrac{1}{n_1^2}-\dfrac{1}{n^2_2}\right]$$
Here, $$\Delta E=(13.6)(3)^2\left[\dfrac{1}{1}-\dfrac{1}{4}\right]$$
$$\Delta E=(13.6)9\times \dfrac{3}{4}=91.8$$eV
So, first excitation energy for ground state in $$Li^{++}$$ is $$91.8$$eV
Option B is correct.
Question 8
1 / -0

The average life of an excited state of hydrogen atom is of the order $${ 10 }^{ -8 }s$$. The number of revolution made by an electron when it is in state n= 2 and before it suffers a transition to state n = 1 are

A
$${ 8.23\times 10 }^{ 6 }$$

B
$${ 2.82\times 10 }^{ 6 }$$

C
$${ 22.8\times 10 }^{ 6 }$$

D
$${ 2.28\times 10 }^{ 6 }$$

Solution
Question 9
1 / -0

The wavelength of the first member of the Balmer series in hydrogen spectrum is $$x$$ $$\mathring{A}$$. Then the wavelength (in $$\mathring{A}$$) of the first member of Lyman series in the same spectrum is

A
$$\cfrac{5}{27}x$$

B
$$\cfrac{4}{3}x$$

C
$$\cfrac{27}{5}x$$

D
$$\cfrac{5}{36}x$$

Solution

$$ \begin{array}{l} \text { Balmer series : transition take place from } \\ n=2 \text { to } n=3,4,5, \text { so on } \\ \text { for first member of balmer } \\ n_{1}=2 \text { to } n_{2}=3 \end{array} $$
$$ \lambda=x \dot {A} $$
$$ \begin{array}{l} \frac{1}{\lambda}=R z^{2}\left(\frac{1}{n_{1} 2}-\frac{1}{n_{2}^{2}}\right) \\ \frac{1}{x}=R z^{2}\left(\frac{1}{4}-\frac{1}{9}\right) \\ \frac{1}{x}=R z^{2}\left(\frac{5}{36}\right)-(1) \end{array} $$
$$ \begin{array}{l} \text { For first member of Lyman } \\ \qquad \begin{aligned} n_{1} &=3 \quad n_{2}=2 \\ \frac{1}{\lambda} &=R_{2}^{2}\left(\frac{1}{1}-\frac{1}{4}\right) \\ \frac{1}{\lambda} &=R z^{2}\left(\frac{3}{4}\right) \end{aligned} \end{array} $$
$$ \begin{array}{l} \text { divide (i) } \div \text { (ii) } \\ \qquad \begin{array}{l} \frac{ \lambda }{x}=\left(\frac{5}{36}\right)\left(\frac{4}{3}\right) \\ \lambda=\left(\frac{5}{27}\right) x \end{array} \end{array} $$
Question 10
1 / -0

In a hypothetical system a particle of mass $$m$$ and charge $$-3q$$ is moving around a very heavy particle having charge $$q$$. Assuming Bohr's model to be true to this system, the orbital velocity of mass $$m$$ when it is nearest to the heavy particle is

A
$$\cfrac{3{q}^{2}}{2{\epsilon}_{0}h}$$

B
$$\cfrac{3{q}^{2}}{4{\epsilon}_{0}h}$$

C
$$\cfrac{3{q}^{}}{2{\epsilon}_{0}h}$$

D
$$\cfrac{3{q}^{}}{4{\epsilon}_{0}h}$$

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Re-Attempt Weekly Quiz Competition

Structure of Atom Test - 72

Result

Structure of Atom Test - 72

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SHARING IS CARING

Question 1

$$He^+$$

$$Li^{++}$$

$$Ne^{9+}$$

$$Na^{10+}$$

Solution

Question 2

$$3.4eV$$

$$13.6eV$$

$$54.4eV$$

$$122.4eV$$

Solution

Question 3

$$49$$

$$50$$

$$51$$

$$25$$

Solution

Question 4

$$28\space g$$

$$14\space g$$

$$7\space g$$

$$20\space g$$

Question 5

$$3x$$

$$9x$$

$$\cfrac{x}{2}$$

$$x$$

Solution

Question 6

n$$(\dfrac{hqB}{8\pi m})$$

n$$(\dfrac{hqB}{4\pi m})$$

n$$(\dfrac{hqB}{2\pi m})$$

n$$(\dfrac{hqB}{\pi m})$$

Solution

Question 7

10.2 eV

91.8 eV

13.6 eV

3.4 eV

Solution

Question 8

$${ 8.23\times 10 }^{ 6 }$$

$${ 2.82\times 10 }^{ 6 }$$

$${ 22.8\times 10 }^{ 6 }$$

$${ 2.28\times 10 }^{ 6 }$$

Solution

Question 9

$$\cfrac{5}{27}x$$

$$\cfrac{4}{3}x$$

$$\cfrac{27}{5}x$$

$$\cfrac{5}{36}x$$

Solution

Question 10

$$\cfrac{3{q}^{2}}{2{\epsilon}_{0}h}$$

$$\cfrac{3{q}^{2}}{4{\epsilon}_{0}h}$$

$$\cfrac{3{q}^{}}{2{\epsilon}_{0}h}$$

$$\cfrac{3{q}^{}}{4{\epsilon}_{0}h}$$

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